Question

the value of p for which one root of the quadratic equation px2-14x+8=0is 6 time the other​

Answer 1

Answer:

let,

q(x)=px^2-14x+8=0 that is a quadratic polynomial

the general form of any quadratic eqn is of the form ax^2+bx+c=0

by comparison we will get,

a=p, b= -14, c=8

let us assume that two zeroes of of q(x) are t and 6t (according to question)

then,

we know

sum of zeroes of any quadratic eqn= -(b)/a

so, t + 6t= -(-14)/p      

       7t=14/p

         t=2/p  __________eqn(1)

also we know that,

product of zeroes=c/a

so, t*6t=8/p

       6t^2=8/p

  putting eqn(1) here we will get

6(2/p)^2=8/p

we will get,

 24/p^2=8/p

24p=8p^2

8p^2-24p=0

p(8p-24)=0

8p-24=0

p=24/8

p=3

hence for value of p=3 q(x) will have roots in which one will be 6 times the other.